124 XIII INTEGRATION
(13.5.11) and the fact that all finite sets are negligible (for Lebesgue meas-
If l = [a, b] is a bounded interval, the "Dirichlet function" (3.11)
cpl — cpl <pQ (which is 0 at the rational points of I and on the complement of
I, and 1 at the irrational points of I) is integrable, and its integral is b — a,
since the set Q is negligible (13.6).
(13.7.9) A subset A ofX. is integrable if and only if, given any B > 0, there
exists a compact set K and an open set G such that KczAcG and
H*(G - K) g 6.
The condition is sufficient by virtue of the definition of integrable functions
and (12.7.4). Suppose conversely that A is integrable, and let e be such that
0 < s < 1. By hypothesis, there exists h e«/ such that (pA^h and
(/z — cpA) d/j,^ e.
Let G be the set of points x e X such that h(x) > 1 — e; then G is open
(12.7.2) and contains A. Clearly h ^> (1 — fi)<po» hence
1 f i
KG)£-----\hdii£-----(/z(A) -f e).
1 -ej 1 -e
Also there exists g e £P such that g ^ <pA and f (<pA — g) d^ g e, and by the
definition of 5^ the set of points x e X such that g(x) > 0 is relatively compact.
Choose 5 > 0 so that §n(A) ^ e, and let K be the set of all x e X such that
g(x) ^ <5; then K is closed in X (12.7.2) and therefore compact (3.17.3), since
it is contained in a relatively compact set. Clearly we have K c: A and
9 =* <PK + ^B > where B = A — K; hence
g dp £ /i(K) + ^(B) ^ XK) + ^(A) ^ ^(K) + e,
and finally //(A) g jg dp 4- e ^ A*(K) + 2e. Q.E.D.
There exist bounded subsets of R which are not integrable with respect to
Lebesgue measure (Section 13.21, Problem 6).
(13.7.10) Let n : X-> X' be a homeomorphism. It follows directly from the
definitions that if/' is any function on X' with values in R, we have
f V • *) dp, f/'dOrGO) = f (/' o n) dfi.a, b is integrable and its measure is \b — a|, by virtue ofion /is integrable if the functions equivalent to /and defined